Homology, Relative Cohomology

Relative Cohomology

If a chain C has subchain A, let the corresponding cochain (C/A)′ be the homomorphisms from cosets of A in C into the reference module M. This is merely the cochain associated with the chain complex C/A, using the implied boundary operator that acts on cosets of A in C. As such, the cochain must indeed be a chain complex, with the corresponding boundary operator running in reverse, and an implied homology. This is the relative cohomology of C/A.

Let f be a chain map from C/A into another chain complex, with A mapping into a subchain of the range. We have seen before that f induces a homomorphism from the relative homology of C/A into the relative homology of the range. Turn f into f′ to carry the relative cochain of the range back to the relative cochain of the domain. This implies a map on relative homologies, which is a reverse map on relative cohomologies.

There are many small theorems one could prove here, about the sum, composition, and restriction of functions having the corresponding effect on relative cohomologies. It all comes out the way you'd expect, and the algebra is the same as before. I'll leave the details to you.

Split Exact

Assume 0 → B → C → C/B → 0 is a split exact sequence of chain complexes. This may seem like an unusual assumption, but many theorems in algebraic topology build a short exact sequence where the third module is free, hence the sequence is split exact.

The functions from C into M determine, and are determined by, the functions from B and from C/B into M. In other words, the cochains 0 ← B′ ← C′ ← (C/B)′ ← 0 are split exact. You may want to take a moment to verify that the left arrows in this sequence, which represent the coboundary operator, follow the reverse projection and reverse injection that are implied by the original split exact sequence. The action of the right arrow, the boundary, does indeed build the left arrow, the coboundary.

Use the split exact sequence of cochains to build a long exact sequence, which is split exact down the middle. This looks like the long exact homology sequence, but the relative cohomology is encountered first at each level, because the arrows run right to left - and you move up the rows instead of down.

Triple Cohomology Sequence

Let the chain C contain the chain B, which contains A. We saw this earlier in the triple homology sequence. Further assume B is a summand of C. Thus the following sequence is split exact.

0 → B/A → C/A → C/B → 0

The cochains into M are split exact, as we demonstrated above, so write the cochains as a short exact sequence and convert to the long exact sequence to get the triple cohomology sequence. Again, arrows run from right to left, and you move up the rows instead of down.

… ← (B/A)′ ← (C/A)′ ← (C/B)′ ← …